gcc編譯c語言中內嵌匯編

--AT&T and Intel 匯編語法對照

寄存器命名：

AT&T:? %eax

Intel: eax

AT&T 語法源地址在左側，目的地址在右側與Intel 方式語法相反

將eax值傳入ebx

AT&T:? movl %eax, %ebx

Intel: mov ebx, eax

AT&T 語法在立即數前有前綴$.

AT&T:? movl $0x0h, %eax

Intel: mov eax,0x0h

AT&T 語法在操作符后跟表示操作數類型的后綴b,w,l分別表示字節，字，雙字，相當于偽操作符ptr，如果不加的話GAS會guess

AT&T:? movw %ax, %bx

Intel: mov bx, ax

內存尋址方式

AT&T:? immed32(basepointer,indexpointer,indexscale)

Intel: [basepointer + indexpointer*indexscale + immed32]

地址計算公式為：

immed32 + basepointer + indexpointer * indexscale

直接尋址

AT&T:? _a

Intel: [_a]

間接尋址

AT&T:? (%eax)

Intel: [eax]

相對尋址

AT&T: _variable(%eax)

Intel: [eax + _variable]

AT&T:? _array(,%eax,4)

Intel: [eax*4 + array]

C 代碼: *(p+1) p定義為char *

AT&T:? 1(%eax) where eax has the value of p

Intel: [eax + 1]

結構體數組尋址，結構體長度為8，下標存于eax，結構體內偏移地址存于ebx，_array為結構體數組首地址

AT&T:? _array(%ebx,%eax,8)

Intel: [ebx + eax*8 + _array]

函數內部實現交換

1、輸入與輸出變量相同

匯編代碼部分標準的交換實現，輸入部分用0寄存器表示"=r"(a)中所指定的寄存器即輸入與輸出變量相同

int main()

{??????????????????????????????????????????????????????? ?　　?????????????????? 804842c:????? mov??? 0xfffffff4(%ebp),%ecx

int a = 10, b = 0;??????????????????????????????????????????????????? 804842f:????? mov??? 0xfffffff0(%ebp),%edx

printf("before swap: a = %2d, b = %2d\n", a , b);?? 8048432:????? mov??? %ecx,%ebx

__asm__("nop;????????????????????????????????????????????????????? 8048434:????? mov??? %edx,%esi

movl %0, %%eax;??????????????????????????????????????? 8048436:????? nop

movl %1, %0;??????????????????????????????????????????????8048437:????? mov??? %ebx,%eax

movl %%eax, %1;??????????????????????????????????????? 8048439:????? mov??? %esi,%ebx

nop;"??????????????????????????????????????????????????????????? 804843b:????? mov??? %eax,%esi

:????????????????????????????????????????????????????????????????????804843d:????? nop

"=r"(a), "=r"(b)???????????????????????????????????????????? 804843e:????? mov??? %ebx,%edx

:??????????????????????????????????????????????????????????????????? 8048440:????? mov??? %esi,%ecx

"0"(a), "1"(b)??????????????????????????????????????????????? 8048442:????? mov??? %edx,%eax

:??????????????????????????????????????????????????????????????????? 8048444:????? mov??? %eax,0xfffffff4(%ebp)

"%eax"?????????????????????????????????????????????????????????8048447:????? mov??? %ecx,%eax

);?????????????????????????????????????????????????????????????????? 8048449:????? mov??? %eax,0xfffffff0(%ebp)

printf("after? swap: a = %2d, b = %2d\n", a, b);

return 0;

}

2、輸入與輸出用不同的寄存器，&表示輸入輸出需要分配不同的寄存器

int main()

{

int a = 10, b = 0;

printf("before swap: a = %2d, b = %2d\n", a, b);????????? 804842b:????? mov??? 0xfffffff8(%ebp),%edx

__asm__("nop;????????????????????????????????????????????????????????????804842e:????? mov??? 0xfffffff4(%ebp),%eax

movl %2, %1;???????????????????????????????????????????????????? 8048431:????? nop

movl %3, %0;?????????????????????????????????????????????????????8048432:????? mov??? %edx,%ebx

nop;"?????????????????????????????????????????????????????????????????? 8048434:????? mov??? %eax,%ecx

:?????????????????????????????????????????????????????????????????????????? 8048436:????? nop

"=&r"(a), "=&r"(b)????????????????????????????????????????????? 8048437:????? mov??? %ecx,%eax

:????????????????????????????????????????????????????????????????????????? ?8048439:????? mov??? %ebx,%edx

"r"(a), "r"(b)??????????????????????????????????????????????????????? 804843b:????? mov??? %eax,%eax

);???????????????????????????????????????????????????????????????????????? 804843d:????? mov??? %eax,0xfffffff8(%ebp)

printf("after? swap: a = %2d, b = %2d\n", a , b);?? ??????? 8048440:????? mov??? %edx,%eax

return 0;?????????????????????????????????????????????????????????????????????? 8048442:????? mov??? %eax,0xfffffff4(%ebp)

}

3、交換函數，需要間接尋址

#include

void swap(int* x, int* y)?????08048400 :

{????????????????????????????????????????????????????????????????????????????????????????? 8048400:????? push?? %ebp

__asm__("nop;???????????????????????????????????????????????????????????? 8048401:????? mov??? %esp,%ebp

movl (%0), %%eax;??????????????????????????????????????????? 8048403:????? push?? %ebx

movl (%1), %%ebx;??????????????????????????????????????????? 8048404:????? mov??? 0x8(%ebp),%ecx

movl %%ebx, (%0);??????????????????????????????????????????? 8048407:????? mov??? 0xc(%ebp),%edx

movl %%eax, (%1);????????????????????????????????????????????804840a:????? nop

nop;"?????????????????????????????????????????????????????????????????? 804840b:????? mov??? (%ecx),%eax

:?????????????????????????????????????????????????????????????????????????? 804840d:????? mov??? (%edx),%ebx

:???????????????????????????????????????????????????????????????????????????804840f:????? mov??? %ebx,(%ecx)

"r"(x),"r"(y)????????????????????????????????????????????????????????? 8048411:????? mov??? %eax,(%edx)

:?????????????????????????????????????????????????????????????????????????? 8048413:????? nop

"eax", "ebx", "memory"??????????????????????????????????????? 8048414:????? mov??? (%esp,1),%ebx ;ebx還原

);?????????????????????????????????????????????????????????????????????????8048417:????? leave?? ;movl %ebp, %esp; pop ebp

}???????????????????????????????????????????????????????????????????????????????????????? 8048418:????? ret

8048419:????? lea??? 0x0(%esi),%esi

int main()

{

int a = 10, b = 0;

printf("before swap: a = %2d, b = %2d\n", a, b);

swap(&a, &b);

printf("after? swap: a = %2d, b = %2d\n", a, b);

return 0;

}

4、從匯編代碼中分離函數

1> 獲得匯編代碼

這里用加法函數，源代碼為：

int sum(int a, int b)

{

int c = a + b;

return c;

}

對應的匯編代碼為

Disassembly of?????????? section .text:

00000000 :

0:?? 55??????????????????????? push?? %ebp

1:?? 89 e5?????????????????? mov??? %esp,%ebp

3:?? 83 ec 04??????????????sub??? $0x4,%esp

6:?? 8b 45 0c??????????????mov??? 0xc(%ebp),%eax

9:?? 03 45 08??????????????add??? 0x8(%ebp),%eax

c:?? 89 45 fc???????????????mov??? %eax,0xfffffffc(%ebp)

f:?? 8b 45 fc????????????????mov??? 0xfffffffc(%ebp),%eax

12:?? 89 c0??????????????????mov??? %eax,%eax

14:?? c9???????????????????????leave

15:?? c3?????????????????????? ret

16:?? 89 f6?????????????????? mov??? %esi,%esi

2> 編寫內嵌匯編語言函數

分析：為函數構建運行時堆棧情況即可使其順利運行，由于編譯器在函數執行開始和結束時會增加

routine begin：

push %ebp; mov %esp, %ebp

routine end:

leave; ret

將上面的0, 1, 14, 15去掉，返回參數放在eax中將輸出部分設置為"=a"(r) 用eax寄存器 r 為需要的return type

步驟：

i?? 定義return_type r 變量

ii? 去掉push %ebp; mov %esp, %ebp;?? leave; ret

iii 輸出部分為:"=a"(r):

$ vi sumassemble.c

int sum(int a, int b)

{

int r;

__asm__("sub $0x4, %%esp;

movl 0xc(%%ebp), %%eax;

addl 0x8(%%ebp), %%eax;

movl %%eax, 0xfffffffc(%%ebp);

movl 0xfffffffc(%%ebp), %%eax;

movl %%eax, %%eax;"

:

"=a"(r)

);

return r;

}

Disassembly of?????????? section .text:

00000000 :

0:?? 55??????????????????????? push?? %ebp

1:?? 89 e5?????????????????? mov??? %esp,%ebp

3:?? 83 ec 04??????????????sub??? $0x4,%esp

6:?? 83 ec 04??????????????sub??? $0x4,%esp

9:?? 8b 45 0c??????????????mov??? 0xc(%ebp),%eax

c:?? 03 45 08??????????????add??? 0x8(%ebp),%eax

f:?? 89 45 fc??????????????? mov??? %eax,0xfffffffc(%ebp)

12:?? 8b 45 fc??????????????mov??? 0xfffffffc(%ebp),%eax

15:?? 89 c0??????????????????mov??? %eax,%eax

17:?? 89 c0??????????????????mov??? %eax,%eax

19:?? 89 45 fc??????????????mov??? %eax,0xfffffffc(%ebp)

1c:?? 8b 45 fc??????????????mov??? 0xfffffffc(%ebp),%eax

1f:?? 89 c0?????????????????? mov??? %eax,%eax

21:?? c9?????????????????????? leave

22:?? c3?????????????????????? ret

23:?? 90?????????????????????? nop

3> 編譯可執行程序

$ vi summain.c

extern int sum(int ,int);

int main()

{

int x = sum(1,2);

printf("x = %d\n", x);

return 0;

}

$ cc -o sum_main sum_main.c sum_assemble.c

$ ./sum_main

總結

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